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Coin rolls on a table

  1. Dec 28, 2009 #1
    1. The problem statement, all variables and given/known data
    if you start a coin rolling on a table with care , u can make it roll in a circle.The coin will lean inward with its axis tilted. The radius of the coin is b.The radius of the circle it follows on the table is R and its velocity is v.assume no slipping.find the angle that axis of coin makes with horizontal.
    (source- klepner kolenkow 7.6)

    2. Relevant equations
    i think they are f(centripetal)=mv^2/R, F(friction)=mew N and torque balance



    3. The attempt at a solution
    mew mg = mv^2/R
    and torque balance gives N R sin theta=F(fric)R cos theta
    gives tan theta=mew=v^2/Rg
    but the answer is given tan theta= 3v^2/2Rg !!!!
     
  2. jcsd
  3. Dec 28, 2009 #2
    Remember that [tex]\vec \tau = \frac{d}{dt}\vec L[/tex]
    Therefore, the net torque cannot be 0! Taking this into account, you should find a different angle. Find the angular momentum vector, and see how it changes in time. Assuming no slipping (No loss of energy), remember that the velocity (Both the linear velocity of the coin and its angular velocity associated with its spin about itself) remains constant in magnitude. This should help you see how the angular momentum vector changes with respect to time.

    Plugging that into the torque equation should give you the result you're looking for.
     
  4. Dec 29, 2009 #3
    thanks royalcat.i understood that a rotating vector is involved here and that is angular momentum. its vertical component remains unchanged but horizontal component is rotating at an angular velocity say W
    so dL=(I (point of contact)* w) cos theta W dt .......(theta=required angle,w= angular vel of spin of coin about its own axis)
    so dL/dt=3 mb^2/2 W w cos theta ......(parallel axis th used)
    this is to be equated to torque about point of contact, right?
    so, mg b sin theta= 3/2 mb^2Ww cos theta
    tan theta=3/2 b Ww /g
    for pure rolling v= bw=RW
    SUBSTITUTING w and W, i get tan theta=3v^2/2Rg.
    was i correct in following you in this? or is there a better method?
    many thanks again.
     
  5. Dec 29, 2009 #4
    That's a very nice way of solving it, my own was much more complicated! Though it could just be because I'm a bit more of a formalist. :)

    Though it is correct mathematically, I think you might want to make a more robust case for your first line:

    [tex]d\vec L = I\omega \cos{\theta} \cdot \Omega dt[/tex]

    I doubt you just pulled it out of nowhere, I'm just curious to see what reasoning you used for it. (Personally I used vectors and geometry to show that [tex]d\vec L = L_{spin}\cos{\theta}\cdot d\phi[/tex] where [tex]d\phi[/tex] is a small angle through which the spin angular momentum vector rotates in the plane of the precession.

    And I then used the fact that [tex]\Omega=\frac{d\phi}{dt}[/tex] and the chain rule to extract an expression for [tex]\frac{d\vec L}{dt}[/tex]

    How did you come up with your expression?
     
  6. Dec 29, 2009 #5
    actually i came up with the formula the same way as u did. like ,say L vector is precessing in a plane so i can resolve the vector as L coswt i + L sin wt j
    so dL= wL dt*(unit vector in tangential direction which is also the direction of torque )
    and hence the result.
    we can also use the triangle rule of vectors to compute dL.
    Personally, i just pulled it from nowhere because we used it many times -in gyros, in precession of mag dipole moment about applied mag field inside an atom etc.:smile:
     
  7. Dec 29, 2009 #6
    Haha, I wish I had your experience, that's a really beautiful argument! I can't wait to meet a question where it's applicable!
     
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